Answer
$\lim\limits_{x \to 3^{+}}\frac{x}{x-3}$ = +$\infty$
Work Step by Step
$\lim\limits_{x \to 3^{+}}\frac{x}{x-3}$ = $\frac{3^{+}}{3^{+}-3}$ = $\frac{3}{0^{+}}$ = +$\infty$
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